Thinking in Waves Episode 63 / Part XX — The Things That Tunnel

Nuclear levels and the Riemann zeros turned out to share a statistic

Waves and zeta ──
line up resonances and you land in number theory “Does treating nuclear physics as waves make the calculation simpler?” ── It does. Split it into resonance (a standing wave) and leakage (tunnelling): the first is a few numbers, the second is mass and charge alone.
★★ “And how do waves relate to zeta?” ── the two questions meet in one place. The spacing statistics of lined-up resonances are the same as those of the Riemann zeros (Montgomery–Dyson 1972).
★★★ Here we compute the Riemann zeros ourselves to check ── \(\chi^2\) is 3906 for Poisson, 495 for GOE, 32.7 for GUE.

Tools needed: Episodes 45, 55, 61, 62Computation: kenshou/zeta.js (10 checks, zero deps)★ interactive

Episode 62 found that d+t’s advantage is a resonance, not the barrier. ★ So how many resonances does it take? And what appears when resonances line up?
★★ The answer ── number theory.

01Nuclei as waves ── splitting resonance from leakage

Cutting nuclear physics in two $$\sigma(E)=\underbrace{\frac{S(E)}{E}}_{\text{nuclear side = resonance}}\times\underbrace{e^{-B_G/\sqrt{E}}}_{\text{outside = leakage}}$$

Episode 62 showed the right factor is fixed by mass and charge alone. What about the left? ── let us fit d+t’s \(S(E)\) with one Breit–Wigner.

\(E\) [keV]\(S(E)\) actualone resonanceratio
1012155155341.278
3021198211100.996
6423061218110.946
10011427122221.070
300120210150.844
fitting window\(E_R\) [keV]\(\Gamma\) [keV]typical error
5–300 keV (wide)48.5106.59.6 %
35–110 keV (peak only)48.891.90.5 %
literature (R-matrix pole)64≈75a different model

The core of §01 ── it does get simpler, but there is a price

★★★ One resonance reproduces 5–300 keV to about 10 %, and the peak region to 0.5 %. d+t’s nuclear physics folds into three numbers (\(E_R\), \(\Gamma\), strength).
★★ But that 48.8 keV is not the literature’s “64 keV resonance”. The 64 keV is an R-matrix pole (with an energy-dependent width); ours is the centre of a constant-width Breit–Wigner fitted to \(S(E)\). Both give the same \(\sigma(E)\), yet the numbers differ by 24 %.
★ Neither is wrong; they are different models ── Episode 55’s “value conditional on a model”. If you want a cross-section, 48.7 keV is correct; if you want the resonance position, use 64 keV.
★★★ So: it does get simpler, but what comes out is a model’s number, not the resonance’s.

02What appears when resonances line up

A nucleus has not one resonance but many — in a heavy nucleus, hundreds spaced a few eV apart near the neutron separation energy. ★ Computing each position is hopeless: all the detail of the nuclear force enters.

★★ But Wigner saw in the 1950s: even when individual positions are beyond reach, the distribution of spacings is not — and it does not depend on the nuclear force at all, only on symmetry.

statistic\(P(s)\)\(\langle s^2\rangle\)fraction \(s<0.3\)which systems
Poisson\(e^{-s}\)2.00000.2592uncorrelated (integrable)
GOE\(\frac{\pi}{2}se^{-\pi s^2/4}\)1.27320.0682time-reversal symmetric ── nuclei
GUE\(\frac{32}{\pi^2}s^2e^{-4s^2/\pi}\)1.17810.0187no time-reversal symmetry

★ Everything hinges on \(s\to0\): if levels were uncorrelated, near-coincident pairs would be the most common (\(e^{-s}\) peaks at \(s=0\)). In real nuclei the distribution falls to zero there ── level repulsion.
★★ Whether it falls as \(s^1\) or \(s^2\) separates GOE from GUE. The Nuclear Data Ensemble is known to follow GOE.

03Computing the Riemann zeros ourselves

Now to number theory. The non-trivial zeros of the Riemann zeta function can be found with the Riemann–Siegel formula.

Riemann–Siegel $$Z(t)=2\sum_{n\le N}\frac{\cos\bigl(\theta(t)-t\ln n\bigr)}{\sqrt n}+\text{correction},\qquad N=\left\lfloor\sqrt{t/2\pi}\right\rfloor$$
#our computationknown valuedifference
114.13719614.1347252.5e−3
221.02437721.0220402.3e−3
532.93256232.9350622.5e−3
1049.77264149.7738321.2e−3

★ It is an asymptotic formula, so low \(t\) costs about \(10^{-3}\); the error falls as \(t^{-3/4}\). For statistics what matters more is that nothing is missed or spurious, which we check against Riemann–von Mangoldt’s \(N(T)\simeq\theta(T)/\pi+1\).

rangezeros found\(N(T)\) predicteddifference
\(t=10\ldots1000\)649648.590.41
\(t=10\ldots5000\)45204520.31−0.31
\(t=10^4\ldots1.2\times10^4\)23772376.630.37
\(t=5\times10^4\ldots5.1\times10^4\)14301431.10−1.10

★★★ Agreement to under one zero ── nothing missed, nothing spurious. Usable for statistics.

04★★★ Unfold, then compare the spacings

The density of zeros grows with \(t\) (\(dN/dt=\ln(t/2\pi)/2\pi\)). To see the statistics the density must be flattened to 1 ── unfolding. Conveniently the transformation is already in hand: \(x_n=\theta(t_n)/\pi\).

value
zeros in \(t=10^4\ldots1.4\times10^4\)4808
raw mean spacing (in \(t\))0.8321
theory \(2\pi/\ln(t/2\pi)\)0.8317
mean after unfolding0.999823

Level statistics of the Riemann zeros ── compute them yourself

10^5
1500
★ Press the button to hunt for zeros of \(Z(t)\) on the spot and take the statistics (a few seconds).
window\(\langle s^2\rangle\)\(s<0.3\)\(\chi^2\) Poisson\(\chi^2\) GOE\(\chi^2\) GUE
\(t=10^4\)1.15830.0221422358141.9
\(t=10^5\)1.16290.0225390649532.7
(GUE)1.17810.0187──────
(GOE)1.27320.0682──────
(Poisson)2.00000.2592──────

What decides it is small \(s\). Looking only at the head of the histogram:

\(s\)★ zeros (measured)PoissonGOEGUE
0.070.0250.9280.1170.018
0.220.1250.7990.3400.154
0.380.3330.6870.5270.381
0.820.9340.4380.7590.928
1.270.7340.2790.5590.665

The core of §04 ── the summit of this episode

★★★ At \(s=0.07\), Poisson predicts 0.928, GOE 0.117, GUE 0.018. The measurement is 0.025.
★★ The zeros refuse to sit close together ── the near-coincident pairs that would be most common if they were uncorrelated are almost absent. That is level repulsion, exactly as in a nucleus.
★★★ And the fall is \(s^2\) (GUE) ── steeper than a nucleus’s \(s^1\) (GOE). \(\chi^2\): Poisson 3906, GOE 495, GUE 32.7 ── GUE beats GOE by a factor 15.
★ Raising \(t\) from \(10^4\) to \(10^5\) improves \(\chi^2\) from 41.9 to 32.7. Convergence is slow, which is why Odlyzko went to \(t=10^{20}\).

05Why they agree ── and what cannot be said

★ The answer is that only symmetry matters. Random-matrix level statistics do not depend on the matrix entries at all ── only on whether the matrix is real symmetric (time-reversal) or Hermitian (none).

★★ So systems with nothing in common land on the same statistic. There is no shared physics between the nuclear force and the zeta function. What is shared is only the structure: many levels lined up under a given symmetry.

what can be saidbasis
d+t’s \(S(E)\) fits one resonance10 % residual (0.5 % on the peak)
★ but \(E_R\) is a model’s numberdifferent from the R-matrix 64 keV
we can compute the zeros ourselvesagrees with \(N(T)\) to under one
★ the spacings are not Poisson\(\chi^2\) differs by orders
★ the spacings are close to GUE\(\langle s^2\rangle\) and the histogram
raising \(t\) moves towards GUE\(10^4\) versus \(10^5\)
★★ what cannot be saidwhy
the zeros are eigenvalues of some operatorHilbert–Pólya ── open
GUE agreement supports the Riemann hypothesisa different question; statistics do not imply it
nuclei and zeros are “the same physics”the same statistic is not the same mechanism
this episode proved anythingnumerical confirmation of known results

The core of §05 ── the distinction that matters most

★★★ “The same statistic” is not “the same physics”. Matching symmetry is enough to produce matching statistics, so the agreement is not evidence of a hidden shared mechanism.
★★ What makes it interesting anyway: that the zeros fall into the symmetry class of a quantum system without time-reversal invariance was not anticipated from the number-theory side. ★ In 1972 Montgomery computed the pair correlation of the zeros, mentioned it to Dyson over tea, and was told at once: “that is GUE.”

06And \(-1/12\) ── it was a spectral zeta all along

Episode 45 produced the Casimir \(-1/12\); Episode 61 verified it is independent of the cutoff across five regulators. ★ Here its origin becomes plain.

A one-dimensional box $$\omega_n=\frac{n\pi c}{a},\qquad E=\frac{\hbar}{2}\sum_n\omega_n=\frac{\hbar\pi c}{2a}\sum_n n=\frac{\hbar\pi c}{2a}\,\zeta(-1)=-\frac{\hbar\pi c}{24a}$$

The core of §06

★★★ \(-1/12\) is the value of the zeta function of a wave spectrum. Not “the value of \(1+2+3+\cdots\)” but a number fixed by how the modes are spaced.
★★ Episode 45 got it by a cutoff, Episode 61 by five regulators agreeing, and this episode gives it a name ── three angles on one thing.
★ And §04 is the same subject ── the statistics of zeros is also a property of how modes are spaced. Zeta functions are the language for writing down the arrangement of waves.

07Can you hear the shape of a drum? ── Weyl’s law

A spectrum carries the shape. Counting the eigenvalues of a rectangular drum (\(a\times b\), fixed edges):

Weyl’s law $$N(\Lambda)\simeq\frac{\text{area}}{4\pi}\Lambda-\frac{\text{perimeter}}{4\pi}\sqrt\Lambda+\cdots$$
\(\Lambda\)counted \(N(\Lambda)\)1st term (area)through 2ndresidual
\(10^4\)531557.0530.01.0
\(10^5\)54865570.45484.91.1
\(10^7\)556183557042.3556186.7−3.7

★★ The first term is area, the second perimeter ── 0.001 % agreement at \(10^7\). The shape is readable from the spectrum.
★★★ That is “can you hear the shape of a drum?”. The answer is “you can hear the area and the perimeter (not the shape itself)”.
★ On the zeta side: the poles and residues of the spectral zeta \(\zeta_\Delta(s)=\sum\lambda_n^{-s}\) correspond to area, perimeter and curvature ── the arrangement of waves and the zeta function are two ways of writing one thing.

Exercises

  1. §01 gave different \(E_R\) for different models. Which Episode 55 point is that?
    Show answer
    “A value conditional on a model”, exactly. When Episode 55 sorted 16 of this series’ numbers, five fell in that class.
    ★ What is new here is a concrete case where two models reproducing the same observable (\(\sigma(E)\)) to the same accuracy return parameters differing by 24 %.
    ★★ Neither is wrong. “48.7 keV” means “the centre of a constant-width BW fitted to \(S(E)\)”; “64 keV” means “the R-matrix pole” ── differently defined quantities.
    ★★★ So the handbook entry is: when you write a parameter, write the model with it. The same shape as Episode 60’s step ⑭, “when you state a variance, state the record length.”
    ★ The observable is model-independent while the parameter is not ── that gap is the line between a measured value and a read-off value.
  2. §04 says raising \(t\) moves towards GUE. Why is the convergence so slow?
    Show answer
    Because the zeros’ statistics carry a correction from the primes.
    ★ The correction to the pair correlation falls like \(1/\ln(t/2\pi)\). ★★ At \(t=10^5\), \(\ln(t/2\pi)=9.7\), so the correction is about 10 %. Getting to 1 % needs \(\ln\approx100\), i.e. \(t\sim e^{100}\approx10^{43}\).
    ★★★ In other words it only improves logarithmically ── Episode 53’s “range can only be bought logarithmically”, again. Even Odlyzko’s \(10^{20}\) gives \(\ln=45\), still at the 2 % level.
    ★ The same reason our \(\chi^2\) only moves from 41.9 to 32.7 ── a tenfold increase in \(t\) takes \(\ln\) from 9.7 only to 12.0.
  3. Nuclei are GOE and the zeros are GUE. What does the difference mean?
    Show answer
    Presence or absence of time-reversal symmetry.
    ★ Nuclei are time-reversal symmetric, so the Hamiltonian can be made real symmetric ── GOE. ★★ The zeros being GUE means that if they were eigenvalues of something, that something would not be time-reversal symmetric.
    ★★★ This is the interesting part and simultaneously the part needing most care. It is not evidence that such an operator exists ── Hilbert–Pólya remains open.
    ★ All that can be said is a conditional hint: if one exists, this is its symmetry class. ★★ In Episode 55’s terms that is an inference under assumption, not a measurement. Nuclei in a magnetic field (time-reversal broken) are known to move to GUE, so the correspondence is concrete ── and still only “the same statistic”.

Summary — line up resonances and you land in number theory

“Does treating nuclear physics as waves make it simpler?” ── Yes.

★★★ \(\sigma(E)=\dfrac{S(E)}{E}\,e^{-B_G/\sqrt E}\).
The right factor is mass and charge alone (Episode 62); the left is one resonance (here).

★ d+t’s \(S(E)\) is reproduced by one resonance to 10 % (0.5 % on the peak) ── the nuclear physics folds into three numbers. ★★ But there is a price: the resulting \(E_R=48.8\) keV is not the literature’s 64 keV (a constant-width BW centre versus an R-matrix pole). The same \(\sigma(E)\), parameters 24 % apart ── a textbook case of Episode 55’s “conditional on a model”.

★★★ And when resonances line up, you get level statistics. Individual positions may be unreachable, but the spacing distribution is fixed by symmetry alone (Wigner).

★★★ That statistic is the same as the statistic of the Riemann zeros.

★ We computed the zeros ourselves to check (Riemann–Siegel, agreeing with \(N(T)\) to under one zero). Mean spacing after unfolding 0.999823; \(\langle s^2\rangle=1.163\) against GUE 1.178, GOE 1.273, Poisson 2.000. ★★ \(\chi^2\): Poisson 3906, GOE 495, GUE 32.7 ── at \(s=0.07\) Poisson predicts 0.928 and the measurement is 0.025. The zeros refuse to sit close together ── exactly what nuclear levels do.

★★ But “the same statistic” is not “the same physics”. Random matrices are fixed by symmetry, so unrelated systems agree. ★ Whether the zeros are eigenvalues of anything (Hilbert–Pólya) remains open, and this episode does not touch it.

★★★ As a bonus, \(-1/12\) is now placed ── it is \(\zeta(-1)\) of a wave spectrum: not the value of \(1+2+3+\cdots\) but a number fixed by how modes are spaced. ★ And as Weyl’s law shows (0.001 % at \(10^7\)), a spectrum carries area and perimeter ── the arrangement of waves and the zeta function are two ways of writing one thing.

This document is Episode 63 of the “Thinking in Waves” series. The computations are in kenshou/zeta.js (zero dependencies, the Riemann zeros computed from scratch, all 10 checks passing). Not discoveries of this episode: the single-level Breit–Wigner formula, the Riemann–Siegel formula, Riemann–von Mangoldt’s \(N(T)\), Wigner’s level statistics and level repulsion, the random-matrix symmetry classes (GOE / GUE), Montgomery’s pair-correlation conjecture and its match with Dyson (1972), Odlyzko’s large-scale numerics, zeta-function regularisation, Weyl’s law, and “can you hear the shape of a drum” ── all well-known; this episode is a numerical confirmation. What it did was ① fit d+t’s \(S(E)\) with one resonance and exhibit the resulting 24 % parameter gap against the R-matrix value as a concrete case of model dependence, ② compute the Riemann zeros and validate the count against \(N(T)\) to confirm they are usable for statistics, ③ compare their spacing distribution against GUE / GOE / Poisson by \(\chi^2\), and ④ re-place \(-1/12\) as a spectral zeta value. Limits: the zeros use the Riemann–Siegel main sum plus the leading correction only, giving \(10^{-3}\)-level errors at low \(t\) (below ≈100). The statistics use only the \(t=10^4\) and \(10^5\) windows, nothing like Odlyzko’s \(10^{20}\) ── the imperfect GUE agreement is not a defect of the computation but the surviving \(1/\ln(t/2\pi)\) correction (Exercise 2). That nuclear levels follow GOE is a literature result; no nuclear data were analysed here. This episode makes no claim about any open problem in number theory (neither the Riemann hypothesis nor Hilbert–Pólya is addressed). ── To print, use your browser’s Print → Save as PDF.

Printing / PDF: ⌘+P (Ctrl+P on Windows). “Show answer” opens the solutions. ★ The compute button does not appear in print.